We know $\(x, y\)$ and $z$ are positive integers greater than 1 such that $\(x y z=231\)$; we need to find the value of $\(x+y+z\)$

As $\(\mathrm{x}, \mathrm{y}\)$ and z are positive integers greater than 1 and $\(\mathrm{xyz}=231=3 \times 7 \times 11\)$, their values, in any order, can be 3, 7 and 11 only.

Hence we get, $\(x+y+z=3+7+11=21\)$, so the answer is (C).